Question

If \[\vec a\] is unit vector and projection of \[\vec x\] is along is $2$ unit and \[(\vec a \times \vec x) + \vec b = \vec x\], then \[\vec x\] is given by ?

Answer 1

Hint:Learn about the vector operations, learn the cross product and dot product of two given vectors. To find the projection of any vector, learn how to evaluate the dot product between two given vectors.

Formula used:
The dot product of any given two vectors are given by,
\[\vec A.\vec B = AB\cos \theta \]
where, \[\vec A\] and \[\vec B\] are the two vectors, \[A\] and \[B\] are the magnitude of the vectors respectively and \[\theta \] is the angle between them.
The projection of a vector, \[\vec A\] along another vector \[\vec B\] is given by, \[\vec A.\hat B\] where \[\hat B\] is the unit vector along \[\vec B\] and is given by,
\[\hat B = \dfrac{{\vec B}}{B}\]
Vector cross product of three vectors is given by,
\[{\mathbf{A \times B \times C = (A}}{\mathbf{.C)B - (A}}{\mathbf{.B)C}}\]
where, \[{\mathbf{A}}\] ,\[{\mathbf{B}}\] and \[{\mathbf{C}}\] are three vectors.

Complete step by step answer:
We have given here the projection of \[\vec x\] along \[\vec a\] is 2 units. Now we know that the projection of any vector along another vector is given by the dot product between them.The mathematical expression is given by, \[\vec A.\hat B\] where \[\hat B\] is the unit vector along \[\vec B\] and is given by, \[\hat B = \dfrac{{\vec B}}{B}\]

So, here we have the unit vector \[\vec a\] and the vector \[\vec x\].So, the projection of \[\vec x\]along \[\vec a\] is, \[\vec a.\vec x\]
The value of which is \[2\]. So, by the question we can write, \[\vec a.\vec x = 2\]…….(i)
Now, also we have given,
\[(\vec a \times \vec x) + \vec b = \vec x\]…..(ii)
Now, cross multiplying \[\vec a\]with the both sides of the equation we have,
\[\vec a \times (\vec a \times \vec x) + \vec a \times \vec b = \vec a \times \vec x\]
\[\Rightarrow (\vec a.\vec x)\vec a - (\vec a.\vec a)\vec x + \vec a \times \vec b = \vec x - \vec b\]............[From equation(ii)]
\[\Rightarrow 2\vec a - \vec x + \vec a \times \vec b = \vec x - \vec b\][From equation(i)]
\[\Rightarrow 2\vec x = 2\vec a + \vec b + (\vec a \times \vec b)\]
\[\therefore \vec x = \dfrac{1}{2}[2\vec a + \vec b + (\vec a \times \vec b)]\]

Hence, the value of \[\vec x\] is \[\dfrac{1}{2}[2\vec a + \vec b + (\vec a \times \vec b)]\].

Note: The vector cross product always gives a vector. The magnitude of the vector \[\vec a\] is one. The simple cross product of any two vector is given by, \[{\mathbf{A \times B}} = AB\sin \theta \hat n\] where, \[\hat n\] is the unit vector perpendicular to both the vector. To solve the problem one must be able to recall the formula for the vector cross product of three vectors which can be easily remembered by recalling the phrase BAC-CAB.